If $2^{a}=3^{b}=6^{c}$ then show that $ c=\frac{a b}{a+b} $
Given: $2^{a}=3^{b}=6^{c}$
To find: Show that $ c=\frac{a b}{a+b} $
Solution:
Let $2^a=3^b=6^c=k$
So,$2^a=d$
⇒$k^{\frac{1}{a}}=2$---------------[i].
and $3^b=d$
⇒$k{\frac{1}{b}}=3$----------------[ii]
.
and also $6^c=k$
⇒$k{\frac{1}{c}}=6$-----------------[iii].
we know that $6=2\times3$
Now substituting , 2,3 and 6
$d^{\frac{1}{a}+\frac{1}{b}} = d^{\frac{1}{c}}$
So, $\frac{1}{c} = \frac{1}{a} + \frac{1}{b}$
⇒$\frac{1}{c} = \frac{b+a}{ ab}$
⇒$c=\frac{ab }{ a+b}$ Hence proved
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